How do I calculate gate current volts?

Discussion in 'General Electronics Chat' started by kenw232, Apr 28, 2015.

  1. kenw232

    Thread Starter Active Member

    May 18, 2009
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    I'm a novice. I'd like to know how to measure and calculate the current & volts which trigger a mosfet or SCR gate in a schematic like the one attached. A signal comes in to the opamp and sends the output high, this switches the transistor Q1 and it triggers the gate of the SCR. Is the voltmeter placed from the Point A to B properly placed to measure the voltage that ends up being applied to the gate? By experiment it reads 880mV.

    How do I calculate the amps and volts so I can cross check it by experiment. For example If I = V/R, then current = 5V/150 = 33mA. Should the ammeter show 33mA being applied to the gate?

    Can the base of Q1 (from the opamp) change the 5V from the power supply? I might not be getting a full 5V to resistor R5 correct?
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    Ohm's Law requires the use of the voltage ACROSS a resistance and the current THROUGH a resistance. You can't just grab a convenient V and a convenient R and use them to get an I that has any meaning.

    Once you fire the SCR in this circuit, what will ever cause it to shut off?

    Your voltage meter is only going to show you the voltage across the ammeter terminals, which should be very low (it would ideally be zero).

    When you inject current into the gate it looks a lot like a transistor base-emitter characteristic with a slightly higher Vbe. A gate-cathode voltage of 800mV is pretty common.
     
  3. kenw232

    Thread Starter Active Member

    May 18, 2009
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    Thats why I asked if 5V/150 = 33mA was correct.

    So how do I measure with a voltmeter the voltage applied to the gate? Do I measure the voltage across R5 (eg. 4V), then subtract that from the supply voltage of 5V, so it would be 1V being applied to the gate?
     
  4. #12

    Expert

    Nov 30, 2010
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    You just did. It's 0.88 volts.

    What happens here is that a positive signal arrives at the op-amp input and the op-amp outputs (Vout) most of 5 volts, depending on which exact op-amp you use. The base to emitter junction of the transistor uses up about 0.6 volts, so the emitter will be Vout - 0.6 volts. From that, you subtract the gate voltage of the SCR (about 0.88 volts). Whatever is left over is divided by 150 ohms to find the gate current of the SCR.
     
  5. WBahn

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    Mar 31, 2012
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    You said in your original post that you had measured it to be 0.88V. How did you get that value?

    If you want to know the gate-cathode voltage that was applied, then measure the voltage between the gate and the cathode. The only reason you wouldn't do this would be because either the measurement is too difficult to make (for instance, if the voltage applied was in the form of a pulse that was too short to get a valid reading) or if the act of making the measurement would disturb the circuit too much.

    If you want to get there by starting with 5V (which is the supply voltage minus the cathode voltage) then you have to subtract off the voltage drops across everything else that is between the gate and the supply. That's not only the voltage across the resistor, but also the voltage across the collector-emitter terminals of the transistor.
     
  6. kenw232

    Thread Starter Active Member

    May 18, 2009
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    It always seems to be .88V. I measure between the gate and cathode with the scope. But its suppose to be higher I think.

    Maybe I'm doing it wrong. I want the gate of my SCR to have applied to it 1.8V @ 38mA say. After the .6V drop from the transistor 2N2222 I get 4.4 Vout. I confirm this with my scope. 4.4V - 1.8V = 2.6V. I need to drop this 2.6V. So 2.6V / .038A = 68.4 Ohms. I put a 68.5 ohm resistor in series between the emitter and SCR gate. I then turn on the transistor and measure current with an ameter just before the gate in series, I see ~40mA. Thats fine. I then measure between the gate and cathode of the SCR with my scope, it says .88V. Why? Should it not be 1.8V? What am I missing?
     
  7. Roderick Young

    Member

    Feb 22, 2015
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    An SCR is really has a PNPN structure inside (you can look it up on wikipedia, or maybe here), so is a current controlled device, more like a diode than anything else at the gate. 0.88 volts sounds fine to me. Does the spec sheet suggest that the voltage should be more?
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    Where does this 1.8V target voltage come from?

    When the transistor is turned on, record the voltage at the transistor base and emitter and at the SCR gate (all relative to the cathode).
     
  9. ScottWang

    Moderator

    Aug 23, 2012
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    If the Vce = 0.8V then the current should be about I=(5V-0.8V)/150=28mA.
    The trigger method is not a good way, you can't treating the Rgk as a real load, you can in series two resistors and the voltage of second one used to trigger the SCR.
     
  10. kenw232

    Thread Starter Active Member

    May 18, 2009
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    The 1.8V is from the SCR datasheet. Its more a less a number I also randomly came up with as a way to learn how to make sure a gate gets a proper voltage applied so I don't burn it out. I measured the base of the transistor at 4.48V, the emitter is 3.76V, and the gate of the SCR is 880mV. The current into the gate of the SCR is measured with an ammeter of about 35mA. It's probably inaccurate.

    So I cannot fix the voltage at 1.8V in this scenario? What I said above still applies and confuses me: "4.48V - 1.8V = 2.6V. I need to drop this 2.6V. So 2.6V / .038A (my current requirement) = 68.4 Ohms. I put a 68.5 ohm resistor in series between the emitter and SCR gate. I then turn on the transistor and measure current with an ameter just before the gate in series, I see ~40mA (or now 35mA). Thats fine. I then measure between the gate and cathode of the SCR with my scope, it says .88V. Why?" Should it not say 1.8V?
     
  11. kenw232

    Thread Starter Active Member

    May 18, 2009
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    I don't have a gate->cathode pulldown resistor Rgk. When I add one it fry's the gate of my SCR due to backemf. I have a difficult high voltage circuit and I cannot mitigate transient spikes. So I'm trying to do this without a pulldown resistor.
     
  12. ScottWang

    Moderator

    Aug 23, 2012
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    Assuming that the Ve =3.76V and using 240Ω and 120 Ω to be the voltage divider and then the Ie about 10.4mA.
    The below attached provided two methods to do the trigger, and I also assuming that the Load is a inductive component, so I in parallel a diode with Load to protecting the SCR from back emf.

    SCR_kenw232_ScottWang.gif
     
  13. #12

    Expert

    Nov 30, 2010
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    The datasheet for the SCR said the guaranteed, worst case, maximum voltage, that will ever be required to fire the gate is 1.8 volts. Why do you want the worst possible gate voltage that was ever made by that company?

    You don't. It isn't broken. You can't fix it.
     
  14. kenw232

    Thread Starter Active Member

    May 18, 2009
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    Thanks, I appreciate it.
     
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